Can someone check my solution to the following question please:
Prove that the alternating group $A_5$ has no subgroup of order 30.
I will be assuming the following result:
Theorem 1: Let $N$ be a subgroup of a group $G$ of index 2. Prove that $N$ is a normal subgroup.
Solution: Suppose $A_5$ has a subgroup group $N$ of order 30. The order of $A_5$ is 60 and by lagrange theorem, $|A_5|/|N|=2$. Hence by the Theorem 1, $N$ is of index 2, and so $N$ is normal. But $A_5$ is a simple group. Hence we arrive at a contradiction.
I try to do a simple proof where I don't have to look into the cycle structures of $A_5$ like the first question from here: question 1 and here: question 1, nor making use of the class equation, which Hungerford's Abstract algebra 3rd edition has not introduced at that edition of the text.
Thank you in advance
Best Answer
As you have the simplicity of $A_5$ at hand, you can prove even more, namely that $A_5$ hasn't got subgroups of order $15$, $20$ and $30$.
In general, let $G$ be a nontrivial finite group and $X_G$ the set of all the proper subgroups of $G$: $$X_G:=\{H\subseteq G\mid H\le G \wedge H\ne G\}.$$ If $G$ is simple, then:
$$[G:H]!\ge|G|, \space\forall H\in X_G \tag 1$$
In fact, for $H\in X_G$, the group $G$ acts by left multiplication on the left quotient set $G/H$, and this action has trivial kernel$^\dagger$. So, $G$ embeds into $S_{[G:H]}$, whence $(1)$.
So, $A_5$ can't have subgroups of order $15$, $20$ and $30$, because $(60/k)!<60$ for $k=15, 20, 30$.
$^\dagger$For $H\in X_G$, the kernel $K:=\bigcap_{g\in G}gHg^{-1}$ is a proper normal subgroup of $G$, and hence $K=\{1\}$ for the simplicity of $G$.